Convexity implies $ \frac{\varphi(c)-\varphi(c-h)}{h} \leq \frac{\varphi(c+h)-\varphi(c)}{h}$?, $h>0 $ Suppose $\varphi $ is a convex function on the real line. I wonder if the following is true? For $h>0 $

$\frac{\varphi(c)-\varphi(c-h)}{h} \leq  \frac{\varphi(c+h)-\varphi(c)}{h}$

This seems like a trivial fact that should follow from convexity if one considers the intuitive meaning of a convex function as having increasing slopes for consecutive points on its graph. 
But I got stuck on which numbers to pick and then use in the convex property. Help would be appreciated!
 A: May be, you could just use Taylor expansions around $h=0$. This would give
$$\text{lhs}=\frac{\varphi(c)-\varphi(c-h)}{h}=\varphi '(c)-\frac{1}{2} h \varphi ''(c)+\frac{1}{6} h^2 \varphi
   ^{(3)}(c)-\frac{1}{24} h^3 \varphi ^{(4)}(c)+O\left(h^4\right)$$
$$\text{rhs}= \frac{\varphi(c+h)-\varphi(c)}{h}=\varphi '(c)+\frac{1}{2} h \varphi ''(c)+\frac{1}{6} h^2 \varphi
   ^{(3)}(c)+\frac{1}{24} h^3 \varphi ^{(4)}(c)+O\left(h^4\right)$$
$$\text{rhs-lhs}=h \varphi ''(c)+\frac{1}{12} h^3 \varphi ^{(4)}(c)+O\left(h^4\right)$$
A: For $h > 0$ we have
$$
\begin{align}
&\frac{\varphi(c)-\varphi(c-h)}{h} \leq  \frac{\varphi(c+h)-\varphi(c)}{h} \\
\iff &\varphi(c)-\varphi(c-h) = \varphi(c+h)-\varphi(c) \\
\iff &\varphi(c) \le \frac 12 \varphi(c-h)+ \frac 12 \varphi(c+h)
\end{align}
$$
which is exactly the convexity condition for the function $\varphi$
$$
 \varphi(\lambda x_1 + (1-\lambda) x_2) \le  \lambda\varphi( x_1) + (1-\lambda) \varphi(x_2)
$$
with $x_1 = c-h$, $x_2 = c+h$, and $\lambda = \frac 12$.
A: By convexity, there is a supporting line at $(c,φ(c))$ with slope $k$, so that
$$
φ(x)\ge φ(c)+k(x-c)
$$
for all $x$ in the domain of $φ$. This means that for $h_1,h_2>0$ you get
$$
\frac{φ(c)-φ(c-h_1)}{h_1}\le k\le \frac{φ(c+h_2)-φ(c)}{h_2}.
$$
